= Solution
For every <orthogonal projector> $P$, positivity gives
$$
\operatorname{Tr}(PX)=\operatorname{Tr}(PQ)-\operatorname{Tr}(PR)
\leq\operatorname{Tr}(PQ)\leq\operatorname{Tr}Q.
$$
These inequalities do not require $P$ to commute with $Q$ or $R$: for example, $\operatorname{Tr}(PR)=\operatorname{Tr}(R^{1/2}PR^{1/2})\geq0$, and apply the same observation to $I-P$ and $Q$.
Let $P_+$ project onto the positive spectral subspace of $X$. Then $P_+Q=Q$ and $P_+R=0$, so $\operatorname{Tr}(P_+X)=\operatorname{Tr}Q=D(\rho,\sigma)$. Hence the <variational characterization of trace distance> over projectors is
$$
\boxed{D(\rho,\sigma)=\max_{P=P^\dagger=P^2}\operatorname{Tr}[P(\rho-\sigma)].}
$$
One may include or exclude the zero eigenspace without changing the optimum.
Back to article page
